Buhrs atomic model differed from ruthofords because it explained that electrons exist in specified energy levels surrounding the nucleus. This means that, Ruthoford believed that electrons can't do very much. However, Buhrs' model showed that electrons are much more powerful than anyone else believes they can be.
Answer:
t = 2.2 s
Explanation:
Given that,
Height of the roof, h = 24.15 m
The initial velocity of the pumpkin, u = 0
We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :
![h=ut+\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=h%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Here, u = 0 and a = g
![h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C%5C%5Ct%3D%5Csqrt%7B%5Cdfrac%7B2h%7D%7Bg%7D%7D%20%5C%5C%5C%5Ct%3D%5Csqrt%7B%5Cdfrac%7B2%5Ctimes%2024.15%7D%7B9.8%7D%7D%20%5C%5C%5C%5Ct%3D2.22%5C%20s)
So, it will take 2.22 s for the pumpkin to hit the ground.
Answer:
electricity
If a rod is charged it is because of the electrical force acting on it
Explanation:
The given data is as follows.
![U_{a} = 5.4 \times 10^{-8} J](https://tex.z-dn.net/?f=U_%7Ba%7D%20%3D%205.4%20%5Ctimes%2010%5E%7B-8%7D%20J)
![W_{/text{a to b}} = -1.9 \times 10^{-8} J](https://tex.z-dn.net/?f=W_%7B%2Ftext%7Ba%20to%20b%7D%7D%20%3D%20-1.9%20%5Ctimes%2010%5E%7B-8%7D%20J)
Electric potential energy (
) = ?
Formula to calculate electric potential energy is as follows.
= ![5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)](https://tex.z-dn.net/?f=5.4%20%5Ctimes%2010%5E%7B-8%7D%20J%20-%20%28-1.9%20%5Ctimes%2010%5E%7B-8%7D%20J%29)
= ![7.3 \times 10^{-8} J](https://tex.z-dn.net/?f=7.3%20%5Ctimes%2010%5E%7B-8%7D%20J)
Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is
.
Answer:
Force, |F| = 2100 N
Explanation:
It is given that,
Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s, ![\dfrac{m}{t}=50\ kg/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bm%7D%7Bt%7D%3D50%5C%20kg%2Fs)
Initial speed, v = 42 m/s
The momentum is reduced to zero, final speed, v = 0
The relation between the force and the momentum is given by :
![F=\dfrac{p}{t}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bp%7D%7Bt%7D)
![F=\dfrac{mv}{t}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bmv%7D%7Bt%7D)
![F=50\ kg/s\times 42\ m/s](https://tex.z-dn.net/?f=F%3D50%5C%20kg%2Fs%5Ctimes%2042%5C%20m%2Fs)
|F| = 2100 N
So, the magnitude of the force exerted on the wall is 2100 N. Hence, this is the required solution.